What is maclaurin 1/(1+x^2),x=2 ?

The answer to maclaurin 1/(1+x^2),x=2 is 1-x^2+x^4-x^6+x^8+\ldots

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maclaurin 1/(1+x^2),x=2

maclaurin

\frac{1}{1 + x ^{2}}, x = 2

1 - x^{2} + x^{4} - x^{6} + x^{8} + \dots

Solution steps

Apply the Maclaurin Formula

= 1 + \frac{\frac{d}{d x} ( \frac{1}{1 + x ^{2}} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( \frac{1}{1 + x ^{2}} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( \frac{1}{1 + x ^{2}} ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 1 + \frac{0}{1 !} x + \frac{- 2}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{24}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{- 720}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{40320}{8 !} x^{8} + \dots

Refine

= 1 - x^{2} + x^{4} - x^{6} + x^{8} + \dots

x \to - 9 - lim (\frac{x + 10}{x + 9})

\frac{\partial}{\partial x} (x^{n} y \frac{1}{x ^{2} + y ^{2}})

\int (1 - x^{3})^{2} d x

\frac{d}{d x} (- 4 e^{x} - sin (x) - 9)

\int π^{2}

What is maclaurin 1/(1+x^2),x=2 ?
The answer to maclaurin 1/(1+x^2),x=2 is 1-x^2+x^4-x^6+x^8+\ldots